>
> it
>
>>>>was
>>>>important enough to merit giving it a name.
>>>>
>>>
>>>It's also wrong. Consider,
>>>
>>>R(a*,b*,c*) where a*, b* and c* are each keys,
>>>
>>>R is not in 6NF because it can be decomposed into
>>>
>>>R1(a*,b*) and
>>>R2(a*,c*)

>>>
>>>even though R doesn't have any non-key attributes!
>>
>>Hi Brian,
>>
>>How should I interpret R(a*, b*, c*)? As a relation with three candidate
>>keys, each over one attribute? Or as a relation with a single candidate
>>key over three attributes?
>>
>>In the former case, R is indeed not in 6NF as I understand it. I guess
>>that Bob was thinking about the primary key only when writing down his
>>explanation for 6NF.

>
> In this case, R is not in 5NF either, is it? While Bob didn't explicitly
> say so isn't it presumed among us that the definition of 6NF includes the
> restriction that it must also be in 5NF?

Bob just gave a simple and informal explanation for why one relation was
not in 6NF. With the formal definition of 6NF, one can decompose R by
arbitrarily choosing any 2 of the 3 combinations of 2 attributes from R.
Received on Mon Dec 17 2007 - 16:45:28 CET